3.35 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(c g+d g x)^3} \, dx\)

Optimal. Leaf size=151 \[ -\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{2 d g^3 (c+d x)^2}+\frac {b^2 B n \log (a+b x)}{2 d g^3 (b c-a d)^2}-\frac {b^2 B n \log (c+d x)}{2 d g^3 (b c-a d)^2}+\frac {b B n}{2 d g^3 (c+d x) (b c-a d)}+\frac {B n}{4 d g^3 (c+d x)^2} \]

[Out]

1/4*B*n/d/g^3/(d*x+c)^2+1/2*b*B*n/d/(-a*d+b*c)/g^3/(d*x+c)+1/2*b^2*B*n*ln(b*x+a)/d/(-a*d+b*c)^2/g^3+1/2*(-A-B*
ln(e*((b*x+a)/(d*x+c))^n))/d/g^3/(d*x+c)^2-1/2*b^2*B*n*ln(d*x+c)/d/(-a*d+b*c)^2/g^3

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Rubi [A]  time = 0.11, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2525, 12, 44} \[ -\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{2 d g^3 (c+d x)^2}+\frac {b^2 B n \log (a+b x)}{2 d g^3 (b c-a d)^2}-\frac {b^2 B n \log (c+d x)}{2 d g^3 (b c-a d)^2}+\frac {b B n}{2 d g^3 (c+d x) (b c-a d)}+\frac {B n}{4 d g^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(c*g + d*g*x)^3,x]

[Out]

(B*n)/(4*d*g^3*(c + d*x)^2) + (b*B*n)/(2*d*(b*c - a*d)*g^3*(c + d*x)) + (b^2*B*n*Log[a + b*x])/(2*d*(b*c - a*d
)^2*g^3) - (A + B*Log[e*((a + b*x)/(c + d*x))^n])/(2*d*g^3*(c + d*x)^2) - (b^2*B*n*Log[c + d*x])/(2*d*(b*c - a
*d)^2*g^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(c g+d g x)^3} \, dx &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 d g^3 (c+d x)^2}+\frac {(B n) \int \frac {b c-a d}{g^2 (a+b x) (c+d x)^3} \, dx}{2 d g}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 d g^3 (c+d x)^2}+\frac {(B (b c-a d) n) \int \frac {1}{(a+b x) (c+d x)^3} \, dx}{2 d g^3}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 d g^3 (c+d x)^2}+\frac {(B (b c-a d) n) \int \left (\frac {b^3}{(b c-a d)^3 (a+b x)}-\frac {d}{(b c-a d) (c+d x)^3}-\frac {b d}{(b c-a d)^2 (c+d x)^2}-\frac {b^2 d}{(b c-a d)^3 (c+d x)}\right ) \, dx}{2 d g^3}\\ &=\frac {B n}{4 d g^3 (c+d x)^2}+\frac {b B n}{2 d (b c-a d) g^3 (c+d x)}+\frac {b^2 B n \log (a+b x)}{2 d (b c-a d)^2 g^3}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 d g^3 (c+d x)^2}-\frac {b^2 B n \log (c+d x)}{2 d (b c-a d)^2 g^3}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 115, normalized size = 0.76 \[ \frac {\frac {B n \left (2 b^2 (c+d x)^2 \log (a+b x)+(b c-a d) (-a d+3 b c+2 b d x)-2 b^2 (c+d x)^2 \log (c+d x)\right )}{(b c-a d)^2}-2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{4 d g^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(c*g + d*g*x)^3,x]

[Out]

(-2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) + (B*n*((b*c - a*d)*(3*b*c - a*d + 2*b*d*x) + 2*b^2*(c + d*x)^2*Log
[a + b*x] - 2*b^2*(c + d*x)^2*Log[c + d*x]))/(b*c - a*d)^2)/(4*d*g^3*(c + d*x)^2)

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fricas [A]  time = 0.97, size = 266, normalized size = 1.76 \[ -\frac {2 \, A b^{2} c^{2} - 4 \, A a b c d + 2 \, A a^{2} d^{2} - 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} n x - {\left (3 \, B b^{2} c^{2} - 4 \, B a b c d + B a^{2} d^{2}\right )} n + 2 \, {\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} \log \relax (e) - 2 \, {\left (B b^{2} d^{2} n x^{2} + 2 \, B b^{2} c d n x + {\left (2 \, B a b c d - B a^{2} d^{2}\right )} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{4 \, {\left ({\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} g^{3} x^{2} + 2 \, {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} g^{3} x + {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )} g^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*g*x+c*g)^3,x, algorithm="fricas")

[Out]

-1/4*(2*A*b^2*c^2 - 4*A*a*b*c*d + 2*A*a^2*d^2 - 2*(B*b^2*c*d - B*a*b*d^2)*n*x - (3*B*b^2*c^2 - 4*B*a*b*c*d + B
*a^2*d^2)*n + 2*(B*b^2*c^2 - 2*B*a*b*c*d + B*a^2*d^2)*log(e) - 2*(B*b^2*d^2*n*x^2 + 2*B*b^2*c*d*n*x + (2*B*a*b
*c*d - B*a^2*d^2)*n)*log((b*x + a)/(d*x + c)))/((b^2*c^2*d^3 - 2*a*b*c*d^4 + a^2*d^5)*g^3*x^2 + 2*(b^2*c^3*d^2
- 2*a*b*c^2*d^3 + a^2*c*d^4)*g^3*x + (b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3)*g^3)

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giac [A]  time = 6.45, size = 203, normalized size = 1.34 \[ \frac {1}{4} \, {\left (2 \, {\left (\frac {2 \, {\left (b x + a\right )} B b n}{{\left (b c g^{3} - a d g^{3}\right )} {\left (d x + c\right )}} - \frac {{\left (b x + a\right )}^{2} B d n}{{\left (b c g^{3} - a d g^{3}\right )} {\left (d x + c\right )}^{2}}\right )} \log \left (\frac {b x + a}{d x + c}\right ) + \frac {{\left (B d n - 2 \, A d - 2 \, B d\right )} {\left (b x + a\right )}^{2}}{{\left (b c g^{3} - a d g^{3}\right )} {\left (d x + c\right )}^{2}} - \frac {4 \, {\left (B b n - A b - B b\right )} {\left (b x + a\right )}}{{\left (b c g^{3} - a d g^{3}\right )} {\left (d x + c\right )}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*g*x+c*g)^3,x, algorithm="giac")

[Out]

1/4*(2*(2*(b*x + a)*B*b*n/((b*c*g^3 - a*d*g^3)*(d*x + c)) - (b*x + a)^2*B*d*n/((b*c*g^3 - a*d*g^3)*(d*x + c)^2
))*log((b*x + a)/(d*x + c)) + (B*d*n - 2*A*d - 2*B*d)*(b*x + a)^2/((b*c*g^3 - a*d*g^3)*(d*x + c)^2) - 4*(B*b*n
- A*b - B*b)*(b*x + a)/((b*c*g^3 - a*d*g^3)*(d*x + c)))*(b*c/(b*c - a*d)^2 - a*d/(b*c - a*d)^2)

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A}{\left (d g x +c g \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*ln(e*((b*x+a)/(d*x+c))^n)+A)/(d*g*x+c*g)^3,x)

[Out]

int((B*ln(e*((b*x+a)/(d*x+c))^n)+A)/(d*g*x+c*g)^3,x)

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maxima [A]  time = 1.36, size = 259, normalized size = 1.72 \[ \frac {1}{4} \, B n {\left (\frac {2 \, b d x + 3 \, b c - a d}{{\left (b c d^{3} - a d^{4}\right )} g^{3} x^{2} + 2 \, {\left (b c^{2} d^{2} - a c d^{3}\right )} g^{3} x + {\left (b c^{3} d - a c^{2} d^{2}\right )} g^{3}} + \frac {2 \, b^{2} \log \left (b x + a\right )}{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} g^{3}} - \frac {2 \, b^{2} \log \left (d x + c\right )}{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} g^{3}}\right )} - \frac {B \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )}{2 \, {\left (d^{3} g^{3} x^{2} + 2 \, c d^{2} g^{3} x + c^{2} d g^{3}\right )}} - \frac {A}{2 \, {\left (d^{3} g^{3} x^{2} + 2 \, c d^{2} g^{3} x + c^{2} d g^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*g*x+c*g)^3,x, algorithm="maxima")

[Out]

1/4*B*n*((2*b*d*x + 3*b*c - a*d)/((b*c*d^3 - a*d^4)*g^3*x^2 + 2*(b*c^2*d^2 - a*c*d^3)*g^3*x + (b*c^3*d - a*c^2
*d^2)*g^3) + 2*b^2*log(b*x + a)/((b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*g^3) - 2*b^2*log(d*x + c)/((b^2*c^2*d - 2
*a*b*c*d^2 + a^2*d^3)*g^3)) - 1/2*B*log(e*(b*x/(d*x + c) + a/(d*x + c))^n)/(d^3*g^3*x^2 + 2*c*d^2*g^3*x + c^2*
d*g^3) - 1/2*A/(d^3*g^3*x^2 + 2*c*d^2*g^3*x + c^2*d*g^3)

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mupad [B]  time = 4.55, size = 221, normalized size = 1.46 \[ \frac {B\,b^2\,n\,\mathrm {atanh}\left (\frac {2\,a^2\,d^3\,g^3-2\,b^2\,c^2\,d\,g^3}{2\,d\,g^3\,{\left (a\,d-b\,c\right )}^2}+\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{d\,g^3\,{\left (a\,d-b\,c\right )}^2}-\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{2\,d\,\left (c^2\,g^3+2\,c\,d\,g^3\,x+d^2\,g^3\,x^2\right )}-\frac {\frac {2\,A\,a\,d-2\,A\,b\,c-B\,a\,d\,n+3\,B\,b\,c\,n}{2\,\left (a\,d-b\,c\right )}+\frac {B\,b\,d\,n\,x}{a\,d-b\,c}}{2\,c^2\,d\,g^3+4\,c\,d^2\,g^3\,x+2\,d^3\,g^3\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*log(e*((a + b*x)/(c + d*x))^n))/(c*g + d*g*x)^3,x)

[Out]

(B*b^2*n*atanh((2*a^2*d^3*g^3 - 2*b^2*c^2*d*g^3)/(2*d*g^3*(a*d - b*c)^2) + (2*b*d*x)/(a*d - b*c)))/(d*g^3*(a*d
- b*c)^2) - (B*log(e*((a + b*x)/(c + d*x))^n))/(2*d*(c^2*g^3 + d^2*g^3*x^2 + 2*c*d*g^3*x)) - ((2*A*a*d - 2*A*
b*c - B*a*d*n + 3*B*b*c*n)/(2*(a*d - b*c)) + (B*b*d*n*x)/(a*d - b*c))/(2*c^2*d*g^3 + 2*d^3*g^3*x^2 + 4*c*d^2*g
^3*x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*ln(e*((b*x+a)/(d*x+c))**n))/(d*g*x+c*g)**3,x)

[Out]

Exception raised: NotImplementedError

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